Trigonometry identity proof

Question

I am working my way through Gelfands trigonometry book. One of the exercises asks to prove the following identity:

$$ \frac{\sin(a)}{1 + \cos(a)} = \frac{1 - \cos(a)}{\sin(a)}$$

I can reduce the identity so that both sides equal 1. But I can't take one side and turn it into the other.

Answer 1

My strategy is take the LHS and multiply by the RHS. Then multiply by the reciprocal of the RHS as follows.

$$\begin{array}{lll} \frac{\sin a}{1+\cos a}&=&\frac{\sin a}{1+\cos a}\cdot\bigg(\frac{1-\cos a}{\sin a}\cdot\frac{\sin a}{1-\cos a}\bigg)\\ &=&\bigg(\frac{\sin a}{1+\cos a}\cdot\frac{\sin a}{1-\cos a}\bigg)\cdot\frac{1-\cos a}{\sin a}\\ &=&\dots \end{array}$$

Answer 2

Try this: Multiply the first member by

$$\frac{1-\cos a}{1-\cos a}$$

(like in rationalization exercises).

Answer 3

Let $s = \sin a, c = \cos a$.

Multiply the LHS by $1 = \frac{1-c}{1-c}$:

$\frac{s(1-c)}{(1+c)(1+c)} = \frac{s(1-c)}{1-c^2} = \frac{s(1-c)}{s^2} = \frac{1-c}{s}$

Answer 4

Cross-multiply the fractions. Then isolate the $1$. The resulting equation is the fundamental relation between the sine and cosine functions.